How To Find Epsnepal Klt Result?

What are good ways to find result of bitwise AND of two numbers (big numbers)?

• for small numbers I can calculate result by converting numbers into binary number e.g. 14&11 = 10 but what if we have big numbers  e.g. 3465 & 97359 = 3081 It's tedious to convert number first into binary and then find result. like we have shortcut for  calculating left shift(<<) and right shift(>>) operation by multiplying and dividing power of 2 e.g. x<<y  is equal to  x*(2^y)  and x>>y  is equal to  x/(2^y) kind of formula or trick I'm looking for.

• Answer:

  I really don't think there can be an analytic formula to be used here, but you can use this iterative procedure which "hides" away the conversion to and from the binary versions of the numbers. Disclaimer: the procedure is not any simpler than actually converting the numbers in binary, in fact it's very analogous. However, it does not feature the binary numbers explicitly. All in all I'm not sure if this is the sort of answer you were looking for.. My intuition would be that there is really nothing simpler than a procedure of the same likeness. EDIT: The following procedure is obsolete, I added an improvement down. It's the same procedure, but with some of the steps which have been collapsed into only one. Let's assume your numbers are x and y. Start by writing on a side all the powers of two until you reach a number bigger than the smallest of your two numbers. So on a side we will have 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096. In the procedure your current power of two will start from 1. Now simply cross your current number of 2 if and only if both x and y are odd numbers. After having done this, divide x and y by two, and round down if necessary, then pass on to the next power of two. Continue until you end all the powers of two. The result will be the sum of all un-crossed powers of two. Here's an example with  x = 3465 and y = 97359. 1) x and y are both odd, thus 1 is not crossed. I set x = 1732. I set y = 48679. 2) x is even, thus 2 is crossed. I set x = 866. I set y = 24339. 3) x is even, thus 4 is crossed. I set x = 433. I set y = 12169. 4) x and y are both odd, thus 8 is not crossed. I set x = 216. I set y = 6084. 5) x and y are both even, thus 16 is crossed. I set x = 108. I set y = 3042. 6) x and y are both even, thus 32 is crossed. I set x = 54. I set y = 1521. 7) x is even, thus 64 is crossed. I set x = 27. I set y = 760. 8) y is even, thus 128 is crossed. I set x = 13. I set y = 380. 9) y is even, thus 256 is crossed. I set x = 6. I set y = 190. 10) x and y are even, thus 512 is crossed. I set x = 3. I set y = 95. 11) x and y are both odd, thus 1024 is not crossed. I set x = 1. I set y = 47. 12) x and y are both odd, thus 2048 is not crossed. I set x = 0. I set y = 23. Now I can stop, because one of the numbers is 0, and all the other powers of 2 are crossed because one between x and y will always be even. Thus x & y = 1 + 8 + 1024 + 2048 = 3081. EDIT: This is the updated procedure. You can make the procedure faster by grouping some of the divides by 2 as follows. This time you don't have to write down all powers of 2 pre-emptively. Just start with p = 1 and t = 0. 1) check if x and y are both odd, if so, t += p. (you will see that after the very first step, you can only continue to check only is y is odd, because x will always be odd.) 2) if x or y are 0 or 1, then return. t is your answer. 2) Divide x by 2 (rounding down each time when necessary) until it is an odd number again. 3) Divide y by 2 (rounding down each time when necessary) the exact number of times as you did x in step number 2. 4) Multiply p by 2 the exact number of times as you divided by two in the previous steps. 5) repeat from 1. Here's an example with  x = 3465 and y = 97359. I initialize p = 1, t = 0. LOOP 1 1) x and y are both odd, t = 0 + p = 1. 2) I don't return yet. 3) I divide x by 2 three times, x = 433. 4) I divide y by 2 three times, y = 12169. 5) I multiply p by 2 three times, p = 8. LOOP 2 1) y is odd, t = 1 + p = 9. 2) I don't return yet. 3) I divide x by 2 four times, x = 27. 4) I divide y by 2 four times, y = 760. 5) I multiply p by 2 four times, p = 128. LOOP 3 1) y is even, t = 9. 2) I don't return yet. 3) I divide x by 2 one time, x = 13. 4) I divide y by 2 one times, y = 380. 5) I multiply p by 2 one time, p = 256. LOOP 4 1) y is even, t = 9. 2) I don't return yet. 3) I divide x by 2 two times, x = 3. 4) I divide y by 2 two times, y = 95. 5) I multiply p by 2 two times, p = 1024. LOOP 5 1) y is odd, t = 9 + p = 1033. 2) I don't return yet. 3) I divide x by 2 one time, x = 1. 4) I divide y by 2 one time, y = 47. 5) I multiply p by 2 one time, p = 2048. LOOP 6 1) y is odd, t = 1033 + p = 3081. 2) x is 1, thus I can return. t = 3081 is the answer.